Let $z = x + yi$ ($x, y \in \mathbb{R}$), then
$$x + yi = a\cos^4 t \cdot i + 2\left(\frac{1}{2} + bi\right)\cos^2 t \cdot \sin^2 t + (1 + ci)\sin^4 t.$$
Separating real and imaginary parts, we get
$$\begin{aligned} x &= \cos^2 t \cdot \sin^2 t + \sin^4 t = \sin^2 t, \\ y &= a(1-x)^2 + 2b(1-x)x + c x^2 \\ &\quad (0 \le x \le 1). \end{aligned}$$

That is, $y = (a+c-2b)x^2 + 2(b-a)x + a$ ($0 \le x \le 1$)
$$a \quad (0 \le x \le 1) \tag{1}$$
Since $A$, $B$, $C$ are non-collinear, $a+c-2b \ne 0$. So Equation (1) is the segment of a parabola (see the diagram). Furthermore, the midpoints of $AB$ and $BC$ are $D(\frac{1}{4}, \frac{a+b}{2})$ and $E(\frac{3}{4}, \frac{b+c}{2})$, respectively. So the equation of line $DE$ is $y = (c-a)x + \frac{1}{4}(3a+2b-c)$. (2)
Solving Equations (1) and (2) simultaneously, we get $(a+c-2b)(x-\frac{1}{2})^2 = 0$. Then $x = \frac{1}{2}$, since $a+c-2b \ne 0$. So the parabola and line $DE$ have one and only one common point $P(\frac{1}{2}, \frac{a+c+2b}{4})$. Notice that $\frac{1}{4} < \frac{1}{2} < \frac{3}{4}$, so point $P$ is on the segment $DE$ and satisfies Equation (1), as required.

We can solve the problem using the method of complex numbers directly. Let $D$, $E$ be the midpoints of $AB$, $CB$, respectively. Then the complex numbers corresponding to $D$, $E$ are $\frac{1}{2}(z_0 + z_1) = \frac{1}{4} + \frac{a+b}{2}i$, $\frac{1}{2}(z_1 + z_2) = \frac{3}{4} + \frac{b+c}{2}i$, respectively. So, complex number $z$ corresponding to a point on the segment $DE$ satisfies
$$z = \lambda\left(\frac{1}{4} + \frac{a+b}{2}i\right) + (1-\lambda)\left(\frac{3}{4} + \frac{b+c}{2}i\right), \quad 0 \le \lambda \le 1.$$

Substitute the above expression into the equation of the curve
$$z = z_0 \cos^4 t + 2z_1 \cos^2 t \cdot \sin^2 t + z_2 \sin^4 t,$$
and separate the real and imaginary parts from both sides to give the
following two equations,
$$\begin{cases} \frac{3}{4} - \frac{\lambda}{2} = \sin^2 t \cos^2 t + \sin^4 t, \\ \frac{1}{2}[\lambda a + b(1 - \lambda)c] = a \cos^4 t + 2b \sin^2 t \cos^2 t + c \sin^4 t. \end{cases}$$
Eliminating $\lambda$ from the equations, we get
$$\begin{aligned} \frac{3}{4}(a-c) + \frac{b+c}{2} &= a\cos^4 t + (2b+a-c)\sin^2 t \cos^2 t + a\sin^4 t \\ &= a(1 - 2\sin^2 t \cos^2 t) + (2b+a-c)\sin^2 t \cos^2 t \\ &= a + (2b-a-c)\sin^2 t \cos^2 t. \end{aligned}$$
Then $(2b-a-c)(\sin^2 t \cos^2 t - \frac{1}{4}) = 0$. Since $A$, $B$, $C$ are non-collinear, we know that $z_1 \neq \frac{1}{2}(z_0 + z_2)$. So $2b-a-c \neq 0$. Then $\sin^2 t \cos^2 t = \sin^2 t(1 - \sin^2 t) = \frac{1}{4}$, that is, $(\sin^2 t - \frac{1}{2})^2 = 0$. Then we have $\frac{3}{4} - \frac{\lambda}{2} = \frac{1}{4} + (\frac{1}{2})^2 = \frac{1}{2}$, so $\lambda = \frac{1}{2} \in [0, 1]$. That means that the curve and the line $DE$ have one and only one common point, and the complex number corresponding to this common point is
$$z = \frac{1}{2} \left( \frac{1}{4} + \frac{a+b}{2}i \right) + \frac{1}{2} \left( \frac{3}{4} + \frac{b+c}{2}i \right) = \frac{1}{2} + \frac{a+c+2b}{4}i.$$